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if P is positive semidefinite, Q is invertible, P and Q are of the same dimension, then is $Q^TPQ$ still positive semidefinite? If so why?

edit: I am aware that sylvester's law of inertia might be useful here but is it possible to show it without using this law?

Sam
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    write $\mathbf y := Q\mathbf x$ then$ \mathbf x^TQ^TPQ\mathbf x\geq 0$ because...? – user8675309 Oct 25 '22 at 16:43
  • it seems you can use corresponding $x$ to create arbitrary $y$, but why this answer https://math.stackexchange.com/questions/68897/symmetric-positive-definite-matrix-question went extra miles (invoking Sylvester's law of inertia, m=n case) to prove it if the proof is so straightforward? – Sam Oct 25 '22 at 16:59
  • @Sam I wouldn't agree that invoking Sylvester's law of inertia is really "going extra miles"; after all, it's saving yourself work by pointing to an existing result. If you really want to understand it "from scratch" however, the top comment's approach is indeed simpler. – Ben Grossmann Oct 25 '22 at 20:09
  • @BenGrossmann sorry just to double check do you think I understand the top comment correctly (from my understanding that 'it seems you can use corresponding x to create arbitrary y', where the corresponding $x = Q^{-1}y$ and since Q is invertible it is always feasible) – Sam Oct 25 '22 at 20:31
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    Yes, I think you understood it correctly. To frame it a bit differently: for any $x$, we must have $$ x^T(Q^TPQ)x = (Qx)^TP(Qx) = y^TPy \geq 0\quad (\text{where } y = Qx). $$ – Ben Grossmann Oct 25 '22 at 20:35
  • @BenGrossmann thank you so much for your confirmation :) – Sam Oct 25 '22 at 20:37

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